Zoj 2432 (longest ascending subsequence)

It's an introduction to lcis. This DP Method has never been described before. Now I have learned it well. DP is really amazing...

First, DP first defines the status DP [I] [J], which indicates the first I of string a and the first J of string B, the length of lcis ending with B [J.

State transition equation:

If (A [I] = B [J]) DP [I] [J] = max (DP [I-1] [k]) + 1; (1 <= k <j)

Else DP [I] [J] = DP [I-1] [J];

Then select the cyclic order to setAlgorithmThe complexity is reduced to N * n.

 

Greatest common increasing subsequence
Time limit: 2 seconds   Memory limit: 65536 KB   Special Judge

You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal 
Possible length.

Sequence S1, S2 ,..., SN of length N is called an increasing subsequence of a sequence A1, A2 ,..., am of length m if there exist 1 <= I1 <I2 <... <in <= m such that SJ = AIJ for all 1 <= j <= n, and SJ <SJ + 1 for all 1 <= j <n. 

Input

Each sequence is described with m-its length (1 <= m <= 500) and M integer numbers AI (-2 ^ 31 <= AI <2 ^ 31) -The sequence itself.

Output

On the first line of the output print l-the length of the greatest common increasing subsequence of both sequences. on the second line print the subsequence itself. if there are several possible answers, output any of them.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. each input block is in the format indicated in the Problem description. there is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1

5
1 4 2 5-12
4
-12 1 2 4

Sample output

2
1 4

Source:  Northeastern Europe 2003, northern subregion

 

# Include <stdio. h> # Include < String . H> # Include <Iostream> Using   Namespace  STD;  # Define N 550 Struct Node {  Int  X, Y;} path [N] [N];  Int  DP [N] [N];  Int  S [N], t [N];  Int  Main (){  Int  T1;  While (Scanf ( "  % D  " , & T1 )! = EOF ){ While (T1 -- ) {Memset (path,  0 , Sizeof  (PATH ));  Int  N, m; scanf (  "  % D  " ,& N );  For ( Int I = 1 ; I <= N; I ++) Scanf (  "  % D  " ,& S [I]); scanf (  "  % D  " ,& M );  For ( Int I = 1 ; I <= m; I ++ ) Scanf (  "  % D " ,& T [I]); memset (DP,  0 , Sizeof  (DP ));  Int MX = 0  ;  For ( Int I = 1 ; I <= N; I ++ ) {MX = 0  ; Int Tx = 0 , Ty = 0  ;  For ( Int J = 1 ; J <= m; j ++ ) {DP [I] [J] = DP [I- 1  ] [J]; path [I] [J]. x = I- 1  ; Path [I] [J]. Y = J; If (S [I]> T [J] & MX <DP [I- 1  ] [J]) {MX = DP [I- 1  ] [J]; TX = I- 1  ; Ty = J ;}  If (S [I] = T [J]) {DP [I] [J] = Mx + 1 ; Path [I] [J]. x = TX; path [I] [J]. Y = Ty ;}} MX =- 1  ;  Int  ID;  For ( Int I = 1 ; I <= m; I ++ )  If (DP [N] [I]>MX) {MX = DP [N] [I]; ID = I ;}  Int  Save [N];  Int CNT = 0  ;  Int  TX, Ty; TX = N; ty = ID;  While (DP [TX] [ty]! = 0 ){  Int  Tmpx, tmpy; tmpx = Path [TX] [ty]. X; tmpy = Path [TX] [ty]. Y;  If (DP [TX] [ty]! = DP [tmpx] [tmpy]) {save [CNT ++] = T [ty];} TX = Tmpx; ty = Tmpy;} printf (  "  % D \ n  " , MX );  For ( Int I = CNT- 1 ; I> = 0 ; I -- ) Printf (  "  % D  "  , Save [I]); printf (  "  \ N  "  );}}  Return  0  ;}